Let $G$ be a discrete countable group acting on a compact, Hausdorff space $X$. Assume that the action is transitive. Namely, $G\cdot x=X$, for all $x\in X$.
Does it follow that $X$ is finite?
I thought that the answer should be yes, as $G\cdot x$ is then a discrete compact space, and so must be finite. However, I am not sure anymore that I can claim that it is a discrete space. I'd appreciate any help.
So generally any topological space is an image of some discrete space. But in this scenario it works. Not because $G$ is discrete but because it is countable. First of all note that $X$ is also countable as an image of a countable set.
Assume that $X$ is not discrete. Then there is a point $x_0\in X$ which is not isolated. Since $G$ acts on $X$ transitively and $x\mapsto gx$ is a homeomorphism then this shows that no point in $X$ is isolated. But a compact Hausdorff space without isolated points has to be uncountable. For the proof see here (plus some discussion regarding related set theoretic axioms, for safety I assume ZFC). Contradiction.
Since $X$ is discrete and compact then it has to be finite.
Note that the assumption about $G$ being discrete is irrelevant.
Since $G$ is countable and $G\cdot x=X$, we have $X$ is countable, compact, Hausdorff. So by Sierpinski-Mazurkiewicz theorem (this seems like an overkill, but I can't think of any simplifications at the moment), $X$ is homeomorphic to $\omega^\alpha\cdot n+1$, where $\alpha+1$ is the Cantor-Bendixon rank of $X$, and $n\geq 1$ is the cardinality of $X^{(\alpha)}$.
Now we need to rule out $\alpha\geq 1$. For such cases, note that $G$ can only map $\omega^\alpha\cdot m$, $m>0$ to another point $\omega^\alpha\cdot m'$, $m'>0$, because every neighbourhood of $\omega^\alpha\cdot m$ contains a homeomorphic copy of $\omega^\alpha+1$. So the action of the homeomorphism group (hence $G$) is not transitive unless $\alpha=0$.
